From 6886111727e91ee5a616d9e1056a6be3447991a1 Mon Sep 17 00:00:00 2001
From: ArthurP <gpan20020205@gmail.com>
Date: Thu, 2 Sep 2021 12:17:45 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200077.=E7=BB=84?=
 =?UTF-8?q?=E5=90=88=E4=BC=98=E5=8C=96.md=20C=E8=AF=AD=E8=A8=80=E7=89=88?=
 =?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0077.组合优化.md | 51 +++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)

diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md
index 171023dd..136ceb34 100644
--- a/problems/0077.组合优化.md
+++ b/problems/0077.组合优化.md
@@ -242,8 +242,59 @@ var combine = function(n, k) {
 };
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
 
+void backtracking(int n, int k,int startIndex) {
+    //当path中元素个数为k个时，我们需要将path数组放入ans二维数组中
+    if(pathTop == k) {
+        //path数组为我们动态申请，若直接将其地址放入二维数组，path数组中的值会随着我们回溯而逐渐变化
+        //因此创建新的数组存储path中的值
+        int* temp = (int*)malloc(sizeof(int) * k);
+        int i;
+        for(i = 0; i < k; i++) {
+            temp[i] = path[i];
+        }
+        ans[ansTop++] = temp;
+        return ;
+    }
 
+    int j;
+    for(j = startIndex; j <= n- (k - pathTop) + 1;j++) {
+        //将当前结点放入path数组
+        path[pathTop++] = j;
+        //进行递归
+        backtracking(n, k, j + 1);
+        //进行回溯，将数组最上层结点弹出
+        pathTop--;
+    }
+}
+
+int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
+    //path数组存储符合条件的结果
+    path = (int*)malloc(sizeof(int) * k);
+    //ans二维数组存储符合条件的结果数组的集合。（数组足够大，避免极端情况）
+    ans = (int**)malloc(sizeof(int*) * 10000);
+    pathTop = ansTop = 0;
+
+    //回溯算法
+    backtracking(n, k, 1);
+    //最后的返回大小为ans数组大小
+    *returnSize = ansTop;
+    //returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度（都为k）
+    *returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
+    int i;
+    for(i = 0; i < *returnSize; i++) {
+        (*returnColumnSizes)[i] = k;
+    }
+    //返回ans二维数组
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

From 8a35955a04594b22d6ea07af06188580a5c6c9f4 Mon Sep 17 00:00:00 2001
From: ArthurP <gpan20020205@gmail.com>
Date: Thu, 2 Sep 2021 12:33:00 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200977.=E6=9C=89?=
 =?UTF-8?q?=E5=BA=8F=E6=95=B0=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9.md=20C?=
 =?UTF-8?q?=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0977.有序数组的平方.md | 31 ++++++++++++++++++++++++++
 1 file changed, 31 insertions(+)

diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md
index 13142853..089884fc 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -270,7 +270,38 @@ def sorted_squares(nums)
 end
 ```
 
+C:
+```c
+int* sortedSquares(int* nums, int numsSize, int* returnSize){
+    //返回的数组大小就是原数组大小
+    *returnSize = numsSize;
+    //创建两个指针，right指向数组最后一位元素，left指向数组第一位元素
+    int right = numsSize - 1;
+    int left = 0;
 
+    //最后要返回的结果数组
+    int* ans = (int*)malloc(sizeof(int) * numsSize);
+    int index;
+    for(index = numsSize - 1; index >= 0; index--) {
+        //左指针指向元素的平方
+        int lSquare = nums[left] * nums[left];
+        //右指针指向元素的平方
+        int rSquare = nums[right] * nums[right];
+        //若左指针指向元素平方比右指针指向元素平方大，将左指针指向元素平方放入结果数组。左指针右移一位
+        if(lSquare > rSquare) {
+            ans[index] = lSquare;
+            left++;
+        } 
+        //若右指针指向元素平方比左指针指向元素平方大，将右指针指向元素平方放入结果数组。右指针左移一位
+        else {
+            ans[index] = rSquare;
+            right--;
+        }
+    }
+    //返回结果数组
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)